# gen chem

what is the pK (=-logK)at 25C for the reaction below?

Cl2(g)+6H2O(l)+5Cu2+(aq)<-->2ClO3-(aq)+12H+(aq)+5Cu(s)

E0=-1.133V

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1. I think I must have Ecell to calculate pK (or concns of all of the reagents).

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2. E cell is not given, just the E0 cell which is -1.133V

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3. That will be
ln K = nFEo>sub>cell/RT
Use 8.314 for R, 298 for T, 96,485 for F. Calculate K, then convert to pK.

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4. ln K = nFEo>sub>cell/RT should read
ln K = nFEocell/RT

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5. what is "n"

and thank you so much for all the help!

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6. Isn't n = 10?
Cl2 ==> 2ClO3^-
zero on the left to +10 (for both Cl atoms) on the right.
5Cu^+2 ==>5Cu(s)
10 on the left (for 5 Cu^+2) and zero on the right.
Total 10 either way.

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7. with n=10, im getting an answer of 0 when I do e^(k)

i got
lnK= (10)(96485)(-1.133)/(8.314)(298)
lnK=-441.228
e^(-441.228)=0

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8. My calculator reads 2.38 x 10^-192.

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9. i tried that, apparently its not the right answer either =/.

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10. The question asks for pK. 2.38 x 10^-192 is K. pK = -log K = -log 2.38 x 10^-192.

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